Hi John, I'm struggling still with [#19](https://forum.azimuthproject.org/discussion/comment/17198/#Comment_17198), because I'm interpreting that if \\(f : (X,u_X) \to (Y,p) \\) is in \\(FinProb\\), then it is measure-preserving, and It would seem that this ruins our freedom to choose \\(p\\) at will.
Your [post](https://johncarlosbaez.wordpress.com/2011/06/02/a-characterization-of-entropy/), around "Suppose we have two finite sets...", says:
> Suppose we have two finite sets with probability measures, say \\((X,p)\\) and \\((Y,q)\\). Then we define a morphism \\(f: (X,p) \to (Y,q)\\) to be a measure-preserving function: in other words, one for which the probability \\(q_j\\) of any point in \\(Y\\) is the sum of the probabilities \\(p_i\\) of the points in \\(X\\) with \\(f(i) = j\\).
In our case Y is the set of labels of the blocks of the partition, the preimages of the partition label singletons are the partition blocks as subsets of X, and, since X is equipped with the uniform distribution, the sum of the probabilities of the elements of the block is simply proportional to the size of the block, so that would force the value of \\(p\\) at the label.